第1行：一個數N表示序列的長度(1 <= N <= 100000)。
第2 - N + 1行：每行1個數，對應數組元素。(0 <= A[i] <= 10^9)

輸出最少需要修改幾個數使得整個數組是嚴格遞增的。

5
1
2
2
3
4

3
位置的考慮也很重要。

 1 #include<iostream>
 2
 
 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 using namespace std;
 9 typedef long long ll;
10 typedef long double ld;
11 typedef pair<int,int> pr;
12 const double pi=acos(-1);
13 #define
 
 rep(i,a,n) for(int i=a;i<=n;i++)
14 #define per(i,n,a) for(int i=n;i>=a;i--)
15 #define Rep(i,u) for(int i=head[u];i;i=Next[i])
16 #define clr(a) memset(a,0,sizeof(a))
17 #define pb push_back
18 #define mp make_pair
19 #define fi first
20 #define sc second
21 #define pq priority_queue
22 #define
 
 pqb priority_queue <int, vector<int>, less<int> >
23 #define pqs priority_queue <int, vector<int>, greater<int> >
24 #define vec vector
25 ld eps=1e-9;
26 ll pp=1000000007;
27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
28 ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
32 ll read(){ ll ans=0; char last=‘ ‘,ch=getchar();
33 while(ch<‘0‘ || ch>‘9‘)last=ch,ch=getchar();
34 while(ch>=‘0‘ && ch<=‘9‘)ans=ans*10+ch-‘0‘,ch=getchar();
35 if(last==‘-‘)ans=-ans; return ans;
36 }
37 const int N=100005;
38 int a[N],b[N];
39 int main(){
40     int n=read(),nu=0;
41     for (int i=1;i<=n;i++) a[i]=read(),a[i]-=i;
42     for (int i=1;i<=n;i++)
43     if (a[i]>=0){
44         int p=upper_bound(b+1,b+nu+1,a[i])-b;
45         if (p>nu) b[++nu]=a[i];
46         else b[p]=a[i];
47     }
48     printf("%d",n-nu);
49     return 0;
50 }

1294 修改數組